The aim is to prove that if A = 0, then A_q = A_r = A_s = A_t = 0. By hypothesis:

0 = 2 A_q cos (18°) + 2 A_r cos (36°) + 2 A_s cos (54°) + 2 A_t cos (72°)

Substitute radical forms for the cosines and multiply both sides by two:

0 = A_q √(10 + √20) + A_r (√5 + 1) + A_s √(10 − √20) + A_t (√5 − 1)

Elimination of radicals. Rearrange the previous equation:

− A_r (√5 + 1) − A_t (√5 − 1) = A_q √(10 + √20) + A_s √(10 − √20)

At the risk of introducing spurious roots, eliminate some radicals by squaring both sides:

A_r² (6 + 2 √5) + A_t² (6 − 2 √5) + 8 A_r A_t = A_q² (10 + √20) + A_s² (10 − √20) + 2 A_q A_s √80

Expand, and simplify the radicals:

6 A_r² + 2 √5 A_r² + 6 A_t² − 2 √5 A_t² + 8 A_r A_t = 10 A_q² + 2 √5 A_q² + 10 A_s² − 2 √5 A_s² + 8 √5 A_q A_s

Move all radicals to the left-hand member, and divide by two:

√5 (A_r² − A_t² − A_q² + A_s² − 4 A_q A_s) = − 3 A_r² − 4 A_r A_t + 5 A_q² − 3 A_t² + 5 A_s²

Because √5 is irrational, but all of A_q, A_r, A_s, A_t are integral, the only hope of solution is when both of these equations are satisfied:

A_r² − A_t² − A_q² + A_s² − 4 A_q A_s = 0
− 3 A_r² − 4 A_r A_t + 5 A_q² − 3 A_t² + 5 A_s² = 0

Equate the two left-hand members and simplify:

A_r² − A_t² − A_q² + A_s² − 4 A_q A_s = − 3 A_r² − 4 A_r A_t + 5 A_q² − 3 A_t² + 5 A_s²
2 A_r² + A_t² + 2 A_r A_t = 3 A_q² + 2 A_s² + 2 A_q A_s

Modulo two. Filter this last equation into modulo two. First, discard any term that is patently even. Then recall that the product of odd numbers is odd, and the product of anything else is even:

A_t² = 3 A_q² [mod 2]
A_t = A_q [mod 2]

Now recall this equation:

A_r² − A_t² − A_q² + A_s² − 4 A_q A_s = 0

which in modulo two becomes:

A_r² + A_t² + A_q² + A_s² = 0 [mod 2]
A_r + A_t + A_q + A_s = 0 [mod 2]

A_t = A_q [mod 2] implies A_t + A_q = 0 [mod 2]. Substitute the latter to obtain A_r + A_s = 0 [mod 2] which implies A_r = A_s [mod 2]. Now there are four cases:

• A_q, A_r, A_s, A_t are all even.
• A_q and A_t are both even, but A_r and A_s are both odd.
• A_q and A_t are both odd, but A_r and A_s are both even.
• A_q, A_r, A_s, A_t are all odd.

Modulo four. Again recall this equation:

A_r² − A_t² − A_q² + A_s² − 4 A_q A_s = 0

which in modulo four becomes:

A_r² − A_t² − A_q² + A_s² = 0 [mod 4]

Consider the case where A_q and A_t are both even, but A_r and A_s are both odd. Recall that the square of an even number is a multiple of four, and the square of an odd number is one more than a multiple of four. The previous equation then becomes:

1 − 0 − 0 + 1 = 0 [mod 4]
2 = 0 [mod 4] -- FALSE

Similarly dismiss the opposite case, where A_q and A_t are both odd and A_r and A_s are both even. Now two cases are left:

• A_q, A_r, A_s, A_t are all even.
• A_q, A_r, A_s, A_t are all odd.

Modulo eight. Note that the square of an odd number is always one more than a multiple of eight. To see this, write the odd number as 2n + 1; its square S becomes 4n² + 4n + 1.

• If n is even, write n = 2m, then S = 16m² + 8m + 1 = 8(2m² + m) + 1.
• If n is odd, write n = 2m + 1, then S = 4(2m + 1)² + 4(2m + 1) + 1 = 16m² + 24m + 9 = 8(2m² + 3m + 1) + 1

Now consider what happens when all of A_q, A_r, A_s, A_t are odd. Recall, yet again, the following equation:

A_r² − A_t² − A_q² + A_s² − 4 A_q A_s = 0

which appears the same in modulo eight:

A_r² − A_t² − A_q² + A_s² − 4 A_q A_s = 0 [mod 8]

From the comments above pertaining to the square of an odd number, substitute 1 for each of the squares:

1 − 1 − 1 + 1 − 4 A_q A_s = 0 [mod 8]
− 4 A_q A_s = 0 [mod 8]

Since A_q and A_s are odd, write A_q = 2B_q + 1 and A_s = 2B_s + 1, where B_q and B_s are integers. Substitute:

− 4 (2B_q + 1) (2B_s + 1) = 0 [mod 8]
16B_qB_s + 8B_q + 8B_s − 4 = 0 [mod 8]

Discarding the obvious multiples of eight leaves:

−4 = 0 [mod 8] -- FALSE

Now one case is left:

• A_q, A_r, A_s, A_t are all even.

Reduction. Since all the As are even, then must exist integer Bs satisfying the equations A_q = 2B_q, A_r = 2B_r, A_s = 2B_s and A_t = 2B_t. From the original hypothesis A = 0, hence 2B = 0, and after scalar division by two, B = 0. This means that B must be a solution in its own right, and all its components are even.

As the existence of solution A necessitates the existence of solution B = A ÷ 2, the existence of solution B implies the existence of a solution C = B ÷ 2 = A ÷ 4. From this result, one must conclude that all the components of A were multiples of four. By the same token, there must also be solution D = C ÷ 2 = B ÷ 4 = A ÷ 8. Now it seems that all the components of A were multiples of eight. Further, solution E = D ÷ 2, which surely exists, calls for the components of A to be multiples of sixteen.

This solution-generating process can continue indefinitely, meaning that each component of A must have infinitely many factors of two, which cannot happen for a finite nonzero integer. Hence, all the components of a solution to A = 0 are themselves zero.